By Lenz’s law, the induced current must cause an upward magnetic field to oppose the loss of flux.. b If the magnetic field is decreasing into the page, the induced magnetic field must o
Trang 129.1: B f NBA,andB i NBAcos37.0B NBA(1cos37.0)
V
5.29
s0600.0
)0.37cos1)(
m25.0)(
m400.0)(
T10.1)(
80
(
)0.37cos1(
B
29.2: a) Before: B NBA(200)(6.010 5 T)(1210 4 m2)
0:after
;mT1044
s040.0
)m102.1)(
T100.6)(
NBA QR
R t
Q IR t
NBA t
b) A credit card reader is a search coil.
c) Data is stored in the charge measured so it is independent of time
29.4: From Exercise (29.3),
.C1016.20
.1280
.6
)m1020.2(T)05.2)(
29.5: From Exercise (29.3),
.T0973.0)
m1020.3)(
120(
)0.450
.60)(
C1056.3(
2 4
d NA dt
)sT102.1()sT012.0(
3 3 4
3 4 4
t
t NA
1013.1600
V0680
Trang 229.7: a) 1 cos 2 2 0 sin 2 for
NAB T
t NAB
dt
d dt
max
T t
T t T
d) From 0t T2,B is getting larger and points in the z direction This gives a
clockwise current looking down the z axis From T2t T,B is getting smaller but still
points in the z direction This gives a counterclockwise current
1
)T4.1(60sin60
dt
d A
dt
dB A
2 1 0 057 s 1 t
)s057.0)(
T4.1)(
60)(sin
12.0)V12.0(10
ln(1 10)0.057s 1tt 40.4s c) B is getting weaker, so the flux is decreasing By Lenz’s law, the induced current must cause an upward magnetic field to oppose the loss of flux Therefore the induced
current must flow counterclockwise as viewed from above.
Trang 329.9: a) c2rand Ar2 so Ac2/4
2)4(B c
B dt
m570.0)(
21)(
isind and I is clockwise.
29.10: According to Faraday’s law (assuming that the area vector points in the positive
z-direction)
)ckwisecounterclo(
V34s
100.2
)m120.0()T5.1(0
)cos
60min1min)(
rev440()m025.0()T060.0)(
m10.0)(
T20.0)(
revrad2)(
srev56)(
500(2
Trang 429.14: NBA t NBA t NBA
dt
d dt
T0750.0)(
120(
V1040.2
29.16: a) If the magnetic field is increasing into the page, the induced magnetic field
must oppose that change and point opposite the external field’s direction, thus requiring a counterclockwise current in the loop
b) If the magnetic field is decreasing into the page, the induced magnetic field must oppose that change and point in the external field’s direction, thus requiring a clockwise current in the loop
c) If the magnetic field is constant, there is no changing flux, and therefore no
induced current in the loop
29.17: a) When the switch is opened, the magnetic field to the right decreases Therefore
the second coil’s induced current produces its own field to the right That means that the
current must pass through the resistor from point a to point b.
b) If coil B is moved closer to coil A, more flux passes through it toward the right
Therefore the induced current must produce its own magnetic field to the left to oppose
the increased flux That means that the current must pass through the resistor from point b
to point a.
c) If the variable resistor R is decreased, then more current flows through coil A, and
so a stronger magnetic field is produced, leading to more flux to the right through coil B
Therefore the induced current must produce its own magnetic field to the left to oppose
the increased flux That means that the current must pass through the resistor from point b
to point a.
Trang 529.18: a) With current passing from a and is increasing the magnetic, field b
becomes stronger to the left, so the induced field points right, and the induced current must flow from right to left through the resistor
b) If the current passes from b , and is decreasing, then there is less magnetic a
field pointing right, so the induced field points right, and the induced current must flow from right to left through the resistor
c) If the current passes from b and is increasing, then there is more magnetic a,field pointing right, so the induced field points left, and the induced current must flow from left to right through the resistor
29.19: a) is B ⊙ and increasing so the flux of the induced current is clockwise.ind b) The current reaches a constant value so is constant B dB dt 0 and there is
Let q be a positive charge in the moving bar The
magnetic force on this charge F v B,
q which
points upward This force pushes the current in a
counterclockwise direction through the circuit.
(ii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase Hence this current must flow in a
counterclockwise sense.
A22.025
V6.5
Trang 629.21: V
C
JC
mNmmC
sNs
mTm
V675
e) b
)m850.0)(
T850.0(
V620
750.0
V620.0
)T800.0)(
m500.0)(
V00.3
to the right
d) Pmech Fv(0.800N)(7.50m s)6.00W
00.650
.1
)V00.3
Trang 729.25: For the loop pulled through the region of magnetic field,
a)
b)
2 2 0
0
R
L vB ILB F
R
vBL I
sm50.4(
V450
dt
d BA dt
d dt
and)
()
.22
2
0 0
dt
dI nr dt
dI r
nA r
m900(cm
b) r1.00cmE3.3910 4 V m
Trang 829.28: a) 12
dt
dB r dt
dB A dt
dB
22
2
1
2 1
dB r dt
dB r
r dt
d r
2
1
2
2 2
2
dB r
R dt
dB r
R dt
d r
29.29: a) The induced electric field lines are concentric circles since they cause the
current to flow in circles
2
m100.02
2
12
12
dB A r dt
d r r
)m100.0
r R
I
2
)00.4)(
A1075.2(
29.30:
nA
r E dt
dI dt
dI nA nIA
dt
d BA dt
d dt
d
0 0
0
)(
)(
)0350.0(2)mV1000.8(
2 1
0
6
s dt
Trang 9E
A is the area of the solenoid.
For a circular path:
dt
di BA r
E2 constant for all circular paths that enclose the solenoid
So W qE2r constant for all paths outside the solenoid
cm
00.7ifJ1014
B B NA t
s0400.0
)A350.0)(
m9000)(
m1000.8)(
12(
4
1 2
smV1076.8(4
A109.12
dt d
Trang 1029.35: a) 55.7A m
)m0400.0(
A280
2 0
i dt
m0200.022
2
0 2
r B
R r
d) Using Ampere’s Law
.T105.3)280.0()m0400.0(
)m0100.0(22
2
0 2
r B
R r
m1050.2
)V120)(
m1000.3()70.4
3
2 4
A CV Q
i K dt
dE j
.mV102.03)
m10(5.00
C100.900A
3 5
5 0
2 4 9 0
q E
A1080
0 2 4 - 3 0
Trang 1129.38: a) 0.15V/m.
m102.1
A)m)(1610
0.2(
2 6 -
m101.2
m100.2
2 6
ρ A
ρI dt
d dt
)A1014.7(2
21
16 0
)A16(22
5 0
29.39:In a superconductor there is no internal magnetic field, and so there is no changing
flux and no induced emf, and no induced electric field
,0)
29.40:Unless some of the regions with resistance completely fill a cross-sectional area of
a long type-II superconducting wire, there will still be no total resistance The regions of
no resistance provide the path for the current Indeed, it will be like two resistors in parellel, where one has zero resistance and the other is non-zero The equivalent
resistance is still zero
29.41: a) For magnetic fields less than the critical field, there is no internal magnetic
field, so:
Inside the superconductor: 0, (0.130 ˆ (1.03 105 A m) ˆ
0 0
B
Outside the superconductor: B B0 (0.130T iˆ, M 0
b) For magnetic fields greater than the critical field, 0 M 0 both inside and outside the superconductor, and BB0 (0.260T iˆ, both inside and outside the superconductor
Trang 1229.42: a) Just under B (threshold of superconducting phase), the magnetic field in the 1material must be zero, and 55 10 Tˆ (4.38 104 A m) ˆ
0
3 0
1
i i
π r
Nπ ) (0.120T)sin (0.9475V)sin(
|
m100150.0,m10
72.1
;:
L R
0400.0()2()500(
is clockwise
Trang 1329.44: a) The large circuit is an RC circuit with a time constant of
s
200F)1020()10
RC Thus, the current as a function of time is
s 20010
c B
c
a ib
dr r
ib
)(1ln22
0 0
A3.7(
(3.0)ln2
m)200.0()10
4(1
ln
Wb 7
di c
a π
b μ dt
d
ε
Thus, the induced current in the small loop is i R 25(0.6000 81m)mV(1.0m) 54A c) The induced current will act to oppose the decrease in flux from the large loop Thus, the induced current flows counterclockwise
d) Three of the wires in the large loop are too far away to make a significant
contribution to the flux in the small loop–as can be seen by comparing the distance c to
the dimensions of the large loop
Trang 1429.45: a)
b)
s50.0
T80.02
)m50.0(2
2
12
12
2
dt
dB r dt
dB r r dt
dB NA r
dt
t BA d R dt
d R R
2 2 2 2 2
R
t A
B R
B
which is the same as part (b)
Trang 1529.47: a)
22
0 2
a a
i BA
di iR dt
di a a
i dt
d iR
0 a R t
e i i(t) i(t)
a
R dt i
a R t(
e i i
ln)2(0.10
m)50.0(ln(0.010)
2
5 0
e) We can ignore the self-induced currents because it takes only a very short time for them to die out
29.48: a) Choose the area vector to point out of the page Since the area and its
orientation to the magnetic field are fixed, we can write the induced emf in the 10 cm radius loop as
])sV(4.00V)
0.20[(
10m)
10.0
dt
dB π
dt
dB A dt
m)10.0(
10s)
0(s)00.2
0 2
4
dt t π
t B t
Thus,
(20.0V)(2.00s) (2.00V s)(2.00s) 0.902Tm
10T)800.0
c) In part (a) the flux has decreased (i.e., it has become more negative) and in part (b)
the flux has increased Both results agree with the expectations of Lenz’s law
Trang 16Consider a narrow strip of width dxand a distance x from
the long wire
The magnetic field of the wire at the strip is B0I 2x.The flux through the strip is
)/()2( 0Ib dx x Bbdx
B
x
dx Ib
d
20
)(2
|
|
)(2
aln2
0
0 0
a r r
Iabv
v a r r
a Ib
dt
dr dt
B B
Trang 1729.50:a) Rotating about the yaxis:
V
0.945m)
10(6.00T)(0.450)
srad0.35
c) Rotating about the zaxis:
V
0.945m)
10(6.00T)(0.450)srad0.35
29.51: From Example 29.4, εω BAsinωt;max ωBA
For N loops, εmax N ω BA
rpm190min)1s(60rad)/2rev(1)srad20(
V120,
m)100.0(T,5.1,400
max
max 2
N
29.52: a) The flux through the coil is given by NBA cos( t ), where N is the number of
turns, B is the strength of the Earth’s magnetic field, and is the angular velocity of the rotating coil Thus, NBA sin( t ), which has a peak amplitude of 0 NBA.Solving for A we obtain
2 5
T)10(8.0turns)(2000rev)/rad(2/60s)min(1min)/rev(30
V0
m)2(0.0650T)
950.0(
2 2
A B t
B
b) Since the flux through the loop is decreasing, the induced current must produce a
field that goes into the page Therefore the current flows from point a through the
resistor to point b
Trang 18i BdA
r
dr iL
22
0 0
L dt
vBL I
mR
L vB m
F m
ILB F
2
2L mR t
v t v mR
L vB m
F dt
B L
FR v
F R
B L v LB R
LB v ILB
Trang 1929.56: The bar will experience a magnetic force due to the induced current in the loop
According to Example 29.6, the induced voltage in the loop has a magnitude BLv which ,opposes the voltage of the battery, Thus, the net current in the loop is I R BLv The acceleration of the bar is sin(90 ) ( )
mR LB BLv m
ILB m F
a) To find mR BLv LB
dt
dv a t
BL v dt mR
LB BLv
)(
s 3.1 t 2
(5.0kg)(0.90
T)1.5(m)(0.8)]
sm(2.0m)(0.8T)(1.5V12[,
Using this v gives 1.2V
29.58: a) According to Example 29.6 the induced emf is BLv(810 5T)
mV
0.1V96)(300
b) For a bullet that travels south, the induced emf is zero
c) In the direction parallel to the velocity the induced emf is zero
Trang 2029.59: From Ampere’s law (Example 28.9), the magnetic field inside the wire, a
distance r from the axis, is ( ) 2 2
0 Ir R r
Consider a small strip of length W and width dr that
is a distance r from the axis of the wire.
The flux through the strip is
dr r R
IW dr
W r B
2)
0 0
2
dr r R
Trang 2129.60: a) (1 3( ) 2( )3).
0
2 0
2 0
B BA
0 0
0
2 0 0 3
0
2 0
2 0
t
πr B t
t t
t dt
d πr B dt
d
ckwise.counterclo
V,5066.0s
010.0
s100.5s
010.0
s100.5s
010.0
)m0420.0(
6
,s100.5atso6
3 2
3 2
0
3 0
2
0 0
2 0 0
t t
t t
πr B
A103.0
V0.0655
3 total
d) Evaluating the emf at t 1.21102 s, using the equations of part (b):
,V6
t t
Iv μ r
dr π
Iv μ dr πr
Iv μ vBdr d
22
2)
b) The magnetic force is strongest at the top end, closest to the current carrying wire
Therefore, the top end, point a, is the higher potential since the force on positive charges
is greatest there, leading to more positives gathering at that end
c) If the single bar was replaced by a rectangular loop, the edges parallel to the wire would have no emf induced, but the edges perpendicular to the wire will have an emf induced, just as in part (b) However, no current will flow because each edge will have its highest potential closest to the current carrying wire It would be like having two batteries
of opposite polarity connected in a loop
29.62: Wire A:vB0 0
Wire C: vBLsin(0.350m s)(0.120T)(0.500m)sin450.0148V
Wire
D: vBL sin (0.350 m s)(0.120 T) 2(0.500 m) sin 45 0.0210 V
Trang 2229.63: a)dε d r BdrεL rBdr ωL B
0
22
1)
.V0.1642
T)(0.650m)
(0.24)sec
rad
b) The potential difference between its ends is the same as the induced emf
c) Zero, since the force acting on each end points toward the center
.V0410.04
part(a) center
b) For a 50 W power dissipation we would require that the resistance be decreased to half the previous value
c) Using the resistance from part (a) and a bar length of 0.20 m
W0.11Ω
0.090
)]
sm2.0m)(
(0.20)T[(0.25)
2 2
R
a vB IaB F R
vBa R
v v t d mR
a B v
v d R
a vB dt
dv m ma F
2
2 ) ( 0
2 2 2
( 0
2 2 2
2
a B
mRv e
a B
mRv x
t d e
v x d
378.0(
)9.36sinˆ36.9m)(cos250
.0(ˆ)mV(0.924ˆ
mV378.0(
Trang 2329.67: At point ,
22
and
dt
dB qr r q qE F dt
dB r dt
dB A dt
left At point b , the field is the same magnitude as at a since they are the same distance
from the center So ,
2 dt
dB qr
F but upward
At point c, there is no force by symmetry arguments: one cannot have one
direction picked out over any other, so the force must be zero
dt d
d E LE da L0,butEda 0soEda L0
abcda E l abBut since we assumed E ab 0, this contradicts Faraday’s law Thus, we can’t have a uniform electric field abruptly drop to zero in a region in which the magnetic field is constant
29.69: At the terminal speed, the upward force F exerted on the loop due to the induced B
current equals the downward force of gravity: F B mg
2 2
4 1
2 2
2 2 T
2 2
2 2 B
164
)2/()4(and
and,
d
s ρ d
s ρ A
ρL R
sd ρ d
s ρ V ρ m
s B
mgR v
mg R
v s B
R v s B IsB F R Bvs I Bvs ε
R R
m m
Trang 2429.70: B l0
d (no currents in the region) Using the figure, let
0
for 0and0for
qd Cdρ
q dρ Adρ
VA AR
V A
0 K0
d
A K A
Aρ Kε
Q e
Aρ K
Q A
Kε
q t
Q K
dt
j d K dt
dE K t
j
K t c
D
)(
)()
0
0 0
0 0
Q
c K
Trang 2529.72: a) 1.96 10 A m
m2300
mV450.0
j D
0 0
2
srad1091.42
d) The two current densities are out of phase by 90 because one has a sine function and the other has a cosine, so the displacement current leads the conduction current by
90
Trang 2629.73: a) τ G r cm m g, summed over each leg,
)90sin(
42)90sin(
42)90sin(
4)0
4)
mgL
sin
IAB
B μ B
.sinsin
R
BA dt
d R
BA dt
d R
BA R
counterclockwise looking to the direction.kˆ
,sinsin2 2 4 2
2 2
A B
12So
R
L B mgL
mL
5
12cos5
R m
L B L
c) The magnetic torque slows down the fall (since it opposes the gravitational torque) d) Some energy is lost through heat from the resistance of the loop
Trang 2729.74: a) For clarity, figure is rotated so B comes out of the page.
r E E
cos(22butcos
dt
dB a dt
dB r dt
dB A dt
d a
2 loop
cos
but2
2 loop
dt
dB a dt
dB a
L dt
dB R
A R
8
)sT0350.0(m)20.0(8
18
ab
But there is potential drop V IR1.7510 4 V, so the potential difference is zero
Trang 2829.75: a)
b) The induced emf on the side ac is zero, because the electric field is always
perpendicular to the line ac
c) To calculate the total emf in the loop,
dt
dB L dt
dB A dt
V1040
37.7
ac
ac IR
V and the point a is at a higher
potential since the current is flowing from atoc
29.76: a) As the bar starts to slide, the flux is decreasing, so the current flows to increase
the flux, which means it flows from a to b
b) The magnetic force on the bar must eventually equal that of gravity
2 2 2
R
B vL vL
R
LB dt
dA B R
LB dt
d R
LB R
LB iLB
2 2
Rmg v
R
B L v mg
B vL vL
R
B dt
dA B R dt
d R R
B L
g Rm R i
90
2 2 2 2
g Rm P
B L
Rmg mg Fv